Find K that requires minimum increments or decrements of array elements to obtain a sequence of increasing powers of K
Last Updated :
14 Jan, 2022
Given an array consisting of N integers, the task is to find the integer K which requires the minimum number of moves to convert the given array to a sequence of powers of K, i.e. {K0, K1, K2, ......., KN - 1}. In each move, increase or decrease an array element by one.
Examples:
Input: arr[] = {1, 2, 3}
Output: 2
Explanation: By incrementing arr[2] by 1 modifies to {1, 2, 4} which is equal to {20, 21, 22}. Therefore, K = 2.
Input: arr[] = {1, 9, 27}
Output: 5
Explanation: Modifying array to {1, 5, 25} requires minimum number of moves. Therefore, K = 5.
Approach: The idea is to check for all the numbers starting from 1 till the power of a number crosses the maximum value i.e. 1010 (assumed). Follow the steps below to solve the problem:
- Check for all the numbers from 1 to x until the value of xN - 1 < max_element of array + move needed to convert in the power of 1.
- Count the moves needed to make an array in the power of that element.
- If the move needed is minimum than the previous value then update the value of K and min moves.
- Print the value of K.
Below is the implementation of the above approach.
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long
// Function to find the value of K such
// that it takes minimum number of moves
// to convert the array in its power
ll findMinCostInteger(vector<ll>& a)
{
// Size of the array
int n = a.size();
// Finding the sum of the array
ll sum = accumulate(a.begin(),
a.end(), 0);
ll K = 0, minmove = LLONG_MAX;
// Find K for which the count
// of moves will become minimum
for (int i = 1;; ++i) {
ll power = 1, count = 0;
for (ll j = 0; j < n; ++j) {
count += abs(a[j] - power);
if (j != n - 1)
power = power * (ll)i;
// If exceeds maximum value
if (power >= 1e10)
break;
}
if (power >= (1e10)
|| power > (ll)(sum - n)
+ a[n - 1])
break;
// Update minimum moves
if (minmove > count) {
minmove = count;
K = i;
}
}
// Print K corresponds to minimum
// number of moves
cout << K;
}
// Driver Code
int main()
{
// Given vector
vector<ll> a = { 1, 9, 27 };
// Function Call
findMinCostInteger(a);
return 0;
}
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the value of K such
// that it takes minimum number of moves
// to convert the array in its power
static void findMinCostInteger(int a[], int n)
{
// Finding the sum of the array
int sum = 0;
for(int i = 0; i < n; i++)
sum += a[i];
int K = 0, minmove = Integer.MAX_VALUE;
// Find K for which the count
// of moves will become minimum
for(int i = 1;; ++i)
{
int power = 1, count = 0;
for(int j = 0; j < n; ++j)
{
count += Math.abs(a[j] - power);
if (j != n - 1)
power = power * i;
// If exceeds maximum value
if (power >= 1e10)
break;
}
if (power >= (1e10) ||
power > (sum - n) + a[n - 1])
break;
// Update minimum moves
if (minmove > count)
{
minmove = count;
K = i;
}
}
// Print K corresponds to minimum
// number of moves
System.out.println(K);
}
// Driver Code
public static void main(String args[])
{
// Given vector
int []a = { 1, 9, 27 };
// Function Call
findMinCostInteger(a, 3);
}
}
// This code is contributed by bgangwar59
# Python3 program for the above approach
import sys
# Function to find the value of K such
# that it takes minimum number of moves
# to convert the array in its power
def findMinCostInteger(a):
# Size of the array
n = len(a)
# Finding the sum of the array
sm = sum(a)
K = 0
minmove = sys.maxsize
# Find K for which the count
# of moves will become minimum
i = 1
while(1):
power = 1
count = 0
for j in range(n):
count += abs(a[j] - power)
if (j != n - 1):
power = power * i
# If exceeds maximum value
if (power >= 1e10):
break
if (power >= (1e10) or
power > (sm - n) + a[n - 1]):
break
# Update minimum moves
if (minmove > count):
minmove = count
K = i
i += 1
# Print K corresponds to minimum
# number of moves
print(K)
# Driver Code
if __name__ == '__main__':
# Given vector
a = [ 1, 9, 27 ]
# Function Call
findMinCostInteger(a)
# This code is contributed by SURENDRA_GANGWAR
// C# program for the
// above approach
using System;
class GFG{
// Function to find the value
// of K such that it takes
// minimum number of moves
// to convert the array in
// its power
static void findMinCostint(int []a,
int n)
{
// Finding the sum of the array
int sum = 0;
for(int i = 0; i < n; i++)
sum += a[i];
int K = 0, minmove = int.MaxValue;
// Find K for which the count
// of moves will become minimum
for(int i = 1;; ++i)
{
int power = 1, count = 0;
for(int j = 0; j < n; ++j)
{
count += Math.Abs(a[j] - power);
if (j != n - 1)
power = power * i;
// If exceeds maximum value
if (power >= 1e10)
break;
}
if (power >= (1e10) ||
power > (sum - n) +
a[n - 1])
break;
// Update minimum moves
if (minmove > count)
{
minmove = count;
K = i;
}
}
// Print K corresponds to
// minimum number of moves
Console.WriteLine(K);
}
// Driver Code
public static void Main(String []args)
{
// Given vector
int []a = {1, 9, 27};
// Function Call
findMinCostint(a, 3);
}
}
// This code is contributed by gauravrajput1
<script>
// Javascript program for the above approach
// Function to find the value of K such
// that it takes minimum number of moves
// to convert the array in its power
function findMinCostLeteger(a, n)
{
// Finding the sum of the array
let sum = 0;
for(let i = 0; i < n; i++)
sum += a[i];
let K = 0, minmove = Number.MAX_VALUE;
// Find K for which the count
// of moves will become minimum
for(let i = 1;; ++i)
{
let power = 1, count = 0;
for(let j = 0; j < n; ++j)
{
count += Math.abs(a[j] - power);
if (j != n - 1)
power = power * i;
// If exceeds maximum value
if (power >= 1e10)
break;
}
if (power >= (1e10) ||
power > (sum - n) + a[n - 1])
break;
// Update minimum moves
if (minmove > count)
{
minmove = count;
K = i;
}
}
// Print K corresponds to minimum
// number of moves
document.write(K);
}
// Driver Code
// Given vector
let a = [ 1, 9, 27 ];
// Function Call
findMinCostLeteger(a, 3);
// This code is contributed by souravghosh0416
</script>
Time Complexity: O(N * max(x))
Auxiliary Space: O(1)
Similar Reads
Minimum increments or decrements required to convert a sorted array into a power sequence Given a sorted array arr[] consisting of N positive integers, the task is to minimize the total number of increments or decrements of each array element required to convert the given array into a power sequence of any arbitrary integer X. A sequence is called a power sequence of any integer X, if an
8 min read
Minimize subarray increments/decrements required to reduce all array elements to 0 Given an array arr[], select any subarray and apply any one of the below operations on each element of the subarray: Increment by oneDecrement by one The task is to print the minimum number of above-mentioned increment/decrement operations required to reduce all array elements to 0. Examples: Input:
5 min read
Minimum count of increment of K size subarrays required to form a given Array Given an array arr[] and an integer K, the task is to find the minimum number of operations required to change an array B of size N containing all zeros such that every element of B is greater than or equal to arr. i.e., arr[i] >= B[i]. In any operation, you can choose a subarray of B of size K a
8 min read
Minimum number of insertions required such that first K natural numbers can be obtained as sum of a subsequence of the array Given an array arr[] consisting of N positive integers and a positive integer K, the task is to find the minimum number of elements that are required to be inserted such that all numbers from the range [1, K] can be obtained as the sum of any subsequence of the array. Examples: Input: arr[] = {1, 3,
8 min read
Minimum increment operations to make the array in increasing order Given an array of size N and X. Find minimum moves required to make the array in increasing order. In each move one can add X to any element in the array. Examples: Input : a = { 1, 3, 3, 2 }, X = 2 Output : 3 Explanation : Modified array is { 1, 3, 5, 6 } Input : a = { 3, 5, 6 }, X = 5 Output : 0 O
6 min read